My physics teacher didn't really make the homework clear at all :/
So it's asking me to find the position, the height and nature... well, uh, the focal length is 6cm and the principal axis is 12cm... the object's height is 4 cm.. and I don't really get what it is trying to ask me, but like I said, it's asking me to find out the position, height and nature?? Of an image I think? I'm really not sure..
If anyone can help solve this, I will reward ten points. Thanks
So it's asking me to find the position, the height and nature... well, uh, the focal length is 6cm and the principal axis is 12cm... the object's height is 4 cm.. and I don't really get what it is trying to ask me, but like I said, it's asking me to find out the position, height and nature?? Of an image I think? I'm really not sure..
If anyone can help solve this, I will reward ten points. Thanks
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It is difficult to figure out the meaning of " principle axis = 12 cm" here.
But as we have to probably find out the position of image, called v, 12 cm must be u(position of object)
and one cannot find position of image only from f (focal length).
Mirror formula: 1/v + 1/u = 1/f
1/v = 1/f -1/u
= 1/6 - 1/(-12) ...........u is negative as object is placed on left side
=1/4
v= + 4cm ..........(position is positive means image is on right side)
Also, height of image/height of object = h2/h1 = -v/u
Here, h1 = 4cm
h2 = -v(h1)/u
= (-1)(4)(4)/(-12)
= + 1.34 cm
As h2 is positive, the image is erect and virtual
But as we have to probably find out the position of image, called v, 12 cm must be u(position of object)
and one cannot find position of image only from f (focal length).
Mirror formula: 1/v + 1/u = 1/f
1/v = 1/f -1/u
= 1/6 - 1/(-12) ...........u is negative as object is placed on left side
=1/4
v= + 4cm ..........(position is positive means image is on right side)
Also, height of image/height of object = h2/h1 = -v/u
Here, h1 = 4cm
h2 = -v(h1)/u
= (-1)(4)(4)/(-12)
= + 1.34 cm
As h2 is positive, the image is erect and virtual